Question

Composite Water Samples The Fairfield County Department of Public Health tests water for the presence of E. coli (Escherichia coli) bacteria. To reduce laboratory costs, water samples from 10 public swimming areas are combined for one test, and further testing is done only if the combined sample tests positive. Based on past results, there is a 0.005 probability of finding E. coli bacteria in a public swimming area. Find the probability that a combined sample from 10 public swimming areas will reveal the presence of E. coli bacteria. Is that probability low enough so that further testing of the individual samples is rarely necessary?

Short answer

The probability that the combined sample will show the presence of E.coli is equal to 0.0489.

As the probability value is low, the water samples do not need to go for individual further testing.

Step-by-step solution

Given information

The probability that a water sample from public swimming area contains E.coli is equal to 0.005.

The number of pools sampled is 10.

Define the event and probability of “at least one”

The probability that an event occurs at least once is one minus the probability that the event does not occur at all. For any given event A, it has the following notation:

$P\left(Aoccurringatleastonce\right)=1-P\left(Anotoccurring\right)$

Compute the probability that at least one pool has E.coli in the combined sample

Let A be the event that a selected water sample has E.coli.

It has the following probability:

$P\left(A\right)=0.005$

Here, $\overline{A}$is the event that a selected water sample does not have E.coli.

It has the following probability:

$$\begin{gathered} P\left(\overline{A}\right)=1-0.005 \\ =0.995 \end{gathered}$$

The probability that out of 10 selected samples, none has E.coli is computed below:

$$\begin{gathered} P\left(\text{none}\text{has}\text{E.coli}\right)=P\left(\overline{A}\right)\times P\left(\overline{A}\right)\times ...\times P\left(\overline{A}\right)\left(10\text{times}\right) \\ ={\left(0.995\right)}^{10} \\ =0.951 \end{gathered}$$

The probability that the combined sample of 10 water samples has E.coli is equal to the probability that at least one of the 10 samples has E.coli. Thus, it is calculated as follows:

$$\begin{gathered} P\left(\text{at}\text{least}\text{one}\text{has}\text{E.coli}\right)=1-P\left(\text{none}\text{has}\text{E.coli}\right) \\ =1-0.9511 \\ =0.0489 \end{gathered}$$

Therefore, the probability that the combined sample will have E.coli is equal to 0.0489.

Interpret the result 

The value of probability can range between 0 and 1, inclusive of both.

An event with a probability lesser than 0.05 can be considered rare or unusual. Thus, the probability of getting a positive result is low for the combined sample.

This implies that further testing of the samples is not necessary.